Unclear differential equation from a thermodynamics textbook

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The discussion centers on a differential equation from a thermodynamics textbook, specifically how to derive the expression involving TdS. Participants clarify that applying the product rule to the term TS leads to the equation d(TS) = TdS + SdT. The focus is on understanding the transformation from the initial equation to the bolded area, which involves recognizing the properties of derivatives. The conversation emphasizes the importance of the product rule in this context. Overall, the derivation illustrates key concepts in thermodynamics related to energy and entropy.
NODARman
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In the thermodynamics textbook there is written: 𝛿𝐴 = 𝑇𝑑𝑆 βˆ’ π‘‘π‘ˆ = 𝑑(𝑇𝑆) βˆ’ 𝑆𝑑𝑇 βˆ’ π‘‘π‘ˆ = βˆ’π‘‘(π‘ˆ βˆ’ 𝑇𝑆) βˆ’ 𝑆𝑑𝑇 = βˆ’π‘‘πΉ βˆ’ 𝑆𝑑𝑇
How did we get the bolded area from TdS? Is that property of derivative, integral, or something else :/
 
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Product rule
 
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Sorry for being so terse. I was posting from my phone.

Apply the product rule to ##TS##. ##d(TS) = ?##.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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